leetcode 105. 从前序与中序遍历序列构造二叉树

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]

返回如下的二叉树：

    3
   / 
  9  20
    /  
   15   7

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* dfs(vector<int>& preorder, vector<int>& inorder, int pre, int inl, int inr){
        if(inl == inr) return new TreeNode(preorder[pre]);
        if(inl > inr) return NULL;
        TreeNode* root = new TreeNode(preorder[pre]);
        int i = inl;
        while(preorder[pre] != inorder[i] && i <= inr) i++;
        root->left = dfs(preorder, inorder, pre+1, inl, i-1);
        root->right = dfs(preorder, inorder, pre+i-inl+1, i+1, inr);
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        int l = preorder.size()-1;
        return dfs(preorder, inorder, 0, 0, l);
    }
};